Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define INF 1<<30

// dist_from_s[u] + dist_from_t[v] = dist_from_s[t] - 2
int main() {
	cin.tie(0); ios::sync_with_stdio(false);

	/* input */
	int N, M; cin >> N >> M;
	int s, t; cin >> s >> t; s--; t--;
	vector<vector<ll>> G(N);
	for (int i = 0; i < M;i++) {
		int x, y; cin >> x >> y; x--; y--;
		G[x].push_back(y);
		G[y].push_back(x);
	}

	/* initialize */
	vector<ll> d_s(N, INF), d_t(N, INF);
	queue<ll> q_s, q_t;
	
	// d_s(t)[n] := distance n from s(t)
	d_s[s] = 0, d_t[t] = 0;
	q_s.push(s), q_t.push(t);

	/* Dijkstra */
	while (!q_s.empty()) {
		int u = q_s.front(); q_s.pop();
		for (int v : G[u]) {
			if (d_s[v] > d_s[u] + 1) {
				d_s[v] = d_s[u] + 1;
				q_s.push(v);
			}
		}
	}

	while (!q_t.empty()) {
		int u = q_t.front(); q_t.pop();
		for (int v : G[u]) {
			if (d_t[v] > d_t[u] + 1) {
				d_t[v] = d_t[u] + 1;
				q_t.push(v);
			}
		}
	}

	/* solve */
	ll ans = 0;

	// dist_is(t)[i] := sum of distance i from s(t)
	vector<ll> dist_is(N + 1, 0);
	vector<ll> dist_it(N + 1, 0);
	for (int i = 0; i < N;i++) {
		if (d_s[i] != INF) {
			dist_is[d_s[i]]++;
		}
		if (d_t[i] != INF) {
			dist_it[d_t[i]]++;
		}
	}

	ll Const = d_s[t] - 2;
	for (int i = 0; i <= Const;i++) {
		ans += (dist_is[i] * dist_it[Const - i]);
	}
	cout << ans << endl;
}